Question

How do you evaluate `\int e ^ { x } \sqrt { 9+ 2e ^ { x } } d x`?

Answer 1

`inte^xsqrt(9+2e^x)dx = (9+2e^x)^(3/2)/3`

Explanation:

`inte^xsqrt(9+2e^x)dx`

This isn't as bad as it looks, actually!

The first thing we can do is recognize that `sqrt(9+2e^x) = (9+2e^x)^(1/2)`

So our original equation turns into

`inte^x(9+2e^x)^(1/2)dx`

U-Substitution seems to be the best way to go.

Our `u` is the inside of the parenthesis.

`u=9+2e^x`

So,

`(du)/(dx)=2e^x`

We have `dx` in our original equation, don't we?
Let's solve for that, then.

`(du)/(dx)=2e^x` becomes `(du)/(2e^x)=dx`

Now that we have `u` and `dx` values, I am going to simplify the original equation.

`inte^x(u)^(1/2)(du)/(2e^x)`

You should immediately see the `e^x` on the top and the bottom. Let's cancel them!

`int(u)^(1/2)(du)/(2)`

We can take the `1/2` outside of the integral to get it out of the

`1/2int(u)^(1/2)(du)`

Now it looks much easier. (I hope)

The power rule tells us how to solve this.

Remember:

`f(x)=u^a`
`f'(x)=au^(a-1)`

Following that, we can solve our equation.

`1/2int(u)^(1/2)(du)`
`1/2(u)^(3/2)*(2/3)`

Simplifying this, our final answer is

`(u)^(3/2)/3`

Right?

Wrong!

Don't forget to substitute the real value for `u`!

`u=9+2e^x`

Our final answer is:

`(9+2e^x)^(3/2)/3`

You can check the answer by finding its derivative.

Answer 2

`inte^xsqrt(9+2e^x)dx=1/3(9+2x)^(3/2)+C`

Explanation:

we can do this by inspection

`inte^xsqrt(9+2e^x)dx`

`=int e^x(9+2x)^(1/2dx`

the outside is a constant multiplied by the derivative of the bracket, suggesting the integral is the bracket to the power +1

try `d/(dx)((9+2e^x)^(3/2))`

by the chain rule we have`=3/2xx2e^x(9+2e)^(1/2)`

`=3e^x(9+2e)^(1/2)`

comparing with the integral we have

`inte^xsqrt(9+2e^x)dx=1/3(9+2e^x)^(3/2)+C`